Question

When one mole of ethylene gas, C2H4, reacts with fluorine gas, hydrogen fluoride and carbon tetrafluoride gases are formed and 2496.7 kJ of heat is given off. What is ΔHf° for CF4(g)?

Answer 1

Answer: The enthalpy of formation of
`CF_4(g)` is -678.82 kJ/mol

Step-by-step explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

`\Delta H^o_(rxn)=\sum [n* \Delta H_f_((product))]-\sum [n* \Delta H_f_((reactant))]`

For the given chemical reaction:

`C_2H_4(g)+6F_2(g)\rightarrow 4HF(g)+2CF_4(g)`

The equation for the enthalpy change of the above reaction is:

`\Delta H_(rxn)=[(4* \Delta H_f_((HF(g))))+(2* \Delta H_f_((CF_4(g))))]-[(1* \Delta H_f_((C_2H_4(g))))+(6* \Delta H_f_((F_2(g))))]`

We are given:

`\Delta H_f_((C_2H_4(g)))=52.26kJ/mol\\\Delta H_f_((HF(g)))=-271.1kJ/mol\\\Delta H_f_((F_2(g)))=0kJ/mol\\\Delta H_(rxn)=-2496.7kJ`

Putting values in above equation, we get:

`-2496.7=[(4* (-271.7))+(2* \Delta H_f_((CF_4(g))))]-[(1* (52.26))+(6* (0))]\\\\\Delta H_f_((CF_4(g)))=-678.82kJ/mol`

Hence, the enthalpy of formation of
`CF_4(g)` is -678.82 kJ/mol